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Question

If $$e_1$$ & $$e_2$$ are eccentricities of hyperbola & its conjugate hyperbola then find $$\frac{1}{e_1^2} + \frac{1}{e_2^2}$$


Solution

Let the hyperbola be $$\dfrac {x^{2}}{a^{2}}-\dfrac {y^{2}}{b^{2}}=1$$

$$\Rightarrow $$conjugate hyperbola is $$\dfrac {x^{2}}{a^{2}}-\dfrac {y^{2}}{b^{2}}=-1$$

$$\Rightarrow e_{1}=\sqrt {\dfrac {a^{2}+b^{2}}{a^{2}}} $$

$$\Rightarrow e_{2}=\sqrt {\dfrac {a^{2}+b^{2}}{b^{2}}} $$

$$\Rightarrow \dfrac {1}{e_{1}^{2}}+\dfrac {1}{e_{2}^{2}}=\dfrac {a^{2}+b^{2}}{a^{2}+b^{2}}=1$$

$$\Rightarrow \dfrac {1}{e_{1}^{2}}+\dfrac {1}{e_{2}^{2}}=1$$

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